Question

solve the following elementary exponential equation.
$9^{-x} = 3$
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4 points
$x = $

Explanation:

Step1: Express 9 as a power of 3

Since \( 9 = 3^2 \), we can rewrite the left - hand side of the equation \( 9^{-x}=3 \) as \( (3^2)^{-x} \).
Using the power - of - a - power rule \((a^m)^n=a^{mn}\), we have \( (3^2)^{-x}=3^{-2x} \). So the equation becomes \( 3^{-2x}=3^{1} \).

Step2: Set the exponents equal

For the exponential equation \( a^m=a^n \) (where \( a>0,a
eq1 \)), we know that \( m = n \).
In our equation \( 3^{-2x}=3^{1} \), since the base \( a = 3>0,a
eq1 \), we can set the exponents equal: \( - 2x=1 \).

Step3: Solve for x

To solve for \( x \), we divide both sides of the equation \( -2x = 1 \) by \( - 2 \).
\( x=\frac{1}{-2}=-\frac{1}{2} \)

Answer:

\(x = -\frac{1}{2}\)